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CCSS.Math:

the graphs of the linear function f of X is equal to MX plus B and the exponential function G of X is equal to a times R to the X where R is greater than zero pass through the points negative 1 comma 9 so this is negative 1 comma 9 right over here and 1 comma 1 both graphs are given below so this very clearly is the linear function it is a line right over here and this right over here is the exponential function so right over here and given the given the fact that this exponential function keeps decreasing as X gets larger and larger and larger it's a pretty good hint that our are right over here they tell us that R is greater than 0 but it's a pretty good hint that R is going to be between 0 and 1 the fact that G of X keeps approaching it's getting closer and closer and closer to 0 as x increases but let's use the data they're giving us the two points of intersection to figure out what the equations of these of these two functions are so first we can tackle the linear function so f of X is equal to M X plus V so they give us two points we could use those points first to figure out the slope so our M R M right over here that's our slope that's our change in Y over change in X the rate and change of the vertical axis with respect to the horizontal axis so let's see between those two points what is our change in X our change in X if we were going from we're going from x equals negative 1 to x equals 1 x equals 1 so we could think of it as we're finishing at 1 we started at negative 1 so 1 minus negative 1 our change in X is 2 we see that right over there and what about our change in Y well we start at 9 let me do this in maybe another color here just we start at 9 and we end up we end up at 1 so we end up at 1 we started at 9 1 minus 9 is negative 8 and just to be clear when X was when X is when X is 1 Y is 1 when X is negative 1 when X is negative 1 Y is 9 another way to think about it the way I drew it right over here we're finishing at x equals 1 y equals 1 we started at x equals negative 1 y equals 9 and so we just took the differences we get negative 8 over 2 which is equal to negative 4 which is equal to negative 4 and so now we can write that f of X f of X is equal to negative 4 negative 4 that's our slope times X negative 4 times X plus B and you could see that slope right over here every every time you increase every time you increase your every time you increase your X by 1 I'm gonna be careful here I got a little bit every time you increase your X by 1 you're decreasing your Y and here on the x axis where we're marking off every half so every time you increase your X by 1 you are decreasing your Y you are decreasing your Y by 4 there so that makes sense that the slope is negative 4 so now let's think about what B is so the figure out B we could use either one of these points to figure out what F of F given an X what f of X is and then we can solve for B so let's say let's try we know let's try F of let's try one because one's a nice simple number so we could write f of 1 which would be negative 4 times 1 plus B and they tell us that F of 1 is 1 is equal to 1 and so this part right over here we could write that as negative 4 plus B is equal to 1 and then we could add 4 to both sides of this equation and then we get B is equal to 5 so we get f of X is equal to negative 4x plus 5 now does that make sense that the y-intercept here is 5 well you see that right over here by inspection you could all you could have you could have guessed actually that the y-intercept here is 5 but now we've and maybe this was five point zero zero zero zero or one or something but now we know for sure it's negative 4x plus five or another way you could have said it if the slope is negative four if this right over here is nine you increase one in the x-direction you're going to decrease four in the y-direction that will get you to Y is equal to five so that is the y-intercept but either way we have figured out the linear function now let's figure out the exponential function so here we could just use the two points to figure out these two unknowns so for example let's just try let's try this first point so G of negative 1 which if we look at this right over here would be a times R to the negative 1 they tell us that G of negative 1 is going to be equal to 9 G of negative 1 is equal to 9 and so we could write we could write this a times R to the negative 1 that's the same thing as a over R is equal to 9 or we could multiply both sides by r and we could say a is equal to 9 r now let's use this other point this other point they tell us they tell us the G of 1 which would be the same thing as a times R to the first power or just a times R that that is equal to that is equal to 1 or a times R is equal to 1 it's not can we use this information right here a is equal to 9 R and a times R is equal to 1 to solve for a and R well I have a little system here it is a nonlinear system but it's a pretty simple one we could just take this a and substitute it in right over here for a and so we would get we would get 9r for a with this first constraint tell us that a must be equal to 9 r so 9 R instead of this instead of writing an 8 here I'll write 9 R times R times R is equal to 1 or we could write or we could write let me scroll down a little bit we could write 9 R squared is equal to 1 divide both sides by 9 R squared is equal to 1 over 9 and now to figure out are you could want it you might want to take the plus the positive and negative square root of both sides but they tell us that R is greater than zero so we can just take the principal root of both sides and we get R is equal to R is equal to one-third and then we could substitute this back into either one of these other two to figure out what a is we know that a is equal to nine times R so nine times one-third a is equal to three so our exponential function could be written as G of X is equal to a which is three times R which is one-third to the 1/3 to the X